Solve the following systems of linear in equations:

$-11 \leq 4 x-3$ and $4 x-3 \leq 13$

When

$-11 \leq 4 x-3$

$4 x-3 \geq-11$

Adding 3 to both the sides

$4 x-3+3 \geq-11+3$

$4 x \geq-8$

Divide both the sides by 4 in above equation

$\frac{4 x}{4} \geq \frac{-8}{4}$

$x \geq-2$

Now when,

$4 x-3 \leq 13$

Adding 3 to both the sides in the above equation

$4 x-3+3 \leq 13+3$

$4 x \leq 16$

Dividing both the sides by 4 in the above question

$\frac{4 x}{4} \leq \frac{16}{4}$

$x \leq 4$

Combining the intervals:

$x \geq-2$ and $x \leq 4$

Therefore,

$x \in[-2,4]$